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eimsori [14]
3 years ago
5

Calculated the gravitational potential energy of a 93.0 kg sky diver who is 550 m above the ground

Physics
1 answer:
Alchen [17]3 years ago
4 0
Equation for Gravitation Potential Energy...
GPe = m·g·h
where m = mass in kg = 93.0 kg
g = force of gravity = 9.81 m/s²
<span>h = height in meters = 550m
</span>
GPe = mgh
GPe = (93)(9.81)(550)
GPe = 501781.5 Joules (J)
GPe = 5.02 x 10⁵ J (rounded & in scientific notation)

Therefore, the gravitational potential energy of a person of mass 93.0kg who is 550m above the ground is GPe = 501781.5 Joules (J)

Hope this helps! 
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ANALOGY, Metal ions: buoys, as electrons: _____.
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Answer:

a. water

Explanation:

A buoy is a floating object that is used in the sea to locate some point or as a checkpoint. It stays at its designated position in the sea by means of an anchor chain. This chain is made short in length according to the water depth do the buoy can not deviate much from its position. The same mechanism can be applied to the metal ion. When a metal ion is formed it remains at its place, but the electrons are mobile and they travel when they get a medium. For example in circuits or from one atom to other. And for the case of buoy, the water serves as electrons as the water is moving in the medium. Hence, the second analogy will be:

electrons : water

So, the correct option is:

<u>a. water</u>

5 0
2 years ago
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3 years ago
How are force and motion related
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5 0
2 years ago
Problem 9: Suppose you wanted to charge an initially uncharged 85 pF capacitor through a 75 MΩ resistor to 90.0% of its final vo
Bingel [31]

Answer:

t=14.678\times 10^{-3}s

Explanation:

Given:

Capacitance, C = 85 pF = 85 × 10⁻¹² F

Resistance, R = 75 MΩ = 75×10⁶Ω

Charge in capacitor at any time 't' is given as:

Q=Q_o(1-e^{-\frac{t}{RC}})

where,

Q₀ = Maximum charge = CE

E = Initial voltage

t = time

also, Q = CV

V= Final voltage = 90% of E = 0.9E

thus, we have

C\times 0.9E=CE(1-e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}})

or

0.9=1-e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}}

or

e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}}=1-0.9=0.1

taking log both sides, we get

ln(e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}})=ln(0.1)=ln(\frac{1}{10})

or

-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}=-ln(10)

or

t=75\times 10^6\ \times\ 85\times 10^{-12}\times ln{10}

or

t=14.678\times 10^{-3}s

3 0
3 years ago
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olga2289 [7]

Answer:

Anticlockwise directions

Please mark me Brainliest to help me

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